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probleem P 1 is reduceerbaar tot P 2 als  afbeelding  :P 1  P 2 zo dat:

algoritme in O(|I| qr ) time voor I in P: doe eerst A  P /* geeft I' in SAT van size O(| | r ) doe dan A. alg. A O(| | q ). yes/no.  P (I)  SAT. A  P O(| | r ). I  P. probleem P 1 is reduceerbaar tot P 2 als  afbeelding  :P 1  P 2 zo dat:

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probleem P 1 is reduceerbaar tot P 2 als  afbeelding  :P 1  P 2 zo dat:

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  1. algoritme in O(|I|qr) time voor I in P: doe eerst AP/* geeft I' in SAT van size O(| |r) doe dan A alg. A O(| |q) yes/no P(I)SAT AP O(| |r) IP probleem P1 is reduceerbaar tot P2 als  afbeelding :P1  P2zo dat: I yes-instantie van P1 (I) yes-instantie van P2 als ook:  polytime-algoritme, zeg A,met inputs IP1 en outputs (I)P2 dan P1 is 'polynomial‑reducable' tot P2 , notatie: P1 P2 TheoremCook (1971): SAT is -compleet, i.e.: each problemP of  is polynomial reducable to SAT Corollary If SAT can be solved by some polynomial algorithm A, then all problems P are polynomial solvable (=)Proof:

  2. YES for k=4 YES for k=3 CLIQUE Instance: graph G=(V, E) ; number k; Question: Is there a ‘clique’ subgraph HG with |V(H)|>=k NO for k=5

  3. Is there a clique of size >= 4 ?

  4. Is there a clique of size >= 4 ?

  5. Satisfiability (formeel) Instance:2n booleans in sets B={b1, b2, ..., bn}en B={b1, b2, ..., bn}, [n-tal B={b1,..., bn} is vrij, n-tal B={b1, b2, ..., bn} is negatie dwz: vrij te kiezen: b1, ..., bn{true,false} en bt = NOT bt voor t=1,2..n] m ‘or-expressies’ C1, C2, ..., Cm met 1 of meer elementen in BB Question:Kan menb1 t/m bn zo kiezen ( truth-assignment)datelke 'clause' C1, C2, ..., Cmtrue wordt. (satisfying all the clauses) bv n=3, m=4 met C1=b1 or b3 or b2 ; C2=b2or b1; C3=b3 or b1; C4=b2 assignment b1=b2=b3=true mislukt; assignment b1=true; b2=b3=false mislukt; assignment b2=true, b1=b3=false satisfies all clauses

  6. elke clause correspondeert met knopenregel, 1 knoop voor elke boolean • alle kanten bestaan behalve: • tussen knopen (booleans) van dezelfde clause • tusen booleans die elkaars negatie zijn Een probleem P zal NP-compleet zijn als: 1) P is in NP 2) voor een bekend NP-compleet P2 geldt: P2  P voorbeeld: we bewijzen Clique is NP-compleet 1) yes-certificaat is subset K met k knopen; O(n2) algoritme checkt kantexistentie 2) we tonen aan SAT CLIQUE : Zij B={b1,…,bn}; C1,C2, …,Cm, een SAT-instantie. bv: B={b1,b2,b3} en m=4 met C1=b1 or b3 or b2 ; C2=b2or b1; C3=b3 or b1; C4=b2

  7. YES:  V’ for k=5 YES:  V’ for k=3 NODE-COVER Instance: graph G=(V, E) ; number k ; Question:can one cover E with a subset V’ of k or less nodes

  8. G’s edges on upper-left node DEFINITION: The co-graphG=(V, co-E)of G=(V, E) has the same node set,but the complement set co-E as edge set: co-E ={(i,j): (i,j)E} Theorem: V’ is a node-cover of E(G) if and only if V\V’ forms a clique in G COROLLARY: CLIQUE  NODE-COVER map I=<(V, E), k> of CLIQUE to (I)=<((V), (E)), (k)> of NODE-COVER, given as (V)=V , (E)=co-E and (k)=|V|-k

  9. CLIQUE NODE-COVERmap I=<(V, E), k> of CLIQUE to (I) of NODECOVER, where  (V,E)=(V,co-E) and (k)=|V|-k to proof:<(V, E), k> is yes instantie clique  <(V,co-E),|V|-k> yes instantie node-cover proof:node set H forms a clique in (V,E) with |H|>=k/* dwz H is yes-certificate for I iff( i, j H: (i,j)  E ) and ( |H|>=k ) iff( i, j H: (i,j)Ïco-E) and ( |H|>=k ) ifffor each edge (i, j) of (V,co-E) node i is in V\H or node j is in V\Hand |V\H|<=|V|-k iff V\H is a node cover of (V, co-E)= (V,E) with |V\H|<=|V|-k/* V\H is yes-certificate for (I), (k)

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